# Force due To Fricton

1. Nov 30, 2011

### xxharulover

I am having a problem with this force equation since it does not state an acceleration but there is a constant velocity. Any help?
Also it does not state any force applied. All i can figure out is normal force and gravitational force.

Here the equation:
A 1500 kg car driving over a dry asphalt surface at a constant speed of 25m/s. There is static friction because you have your anti lock breaks on.

All I know is that the applied force due to constant velocity equals the weight of the car.

From what I know Normal Force Gravitational force is equal so
(9.81 m/s^2)(1500kg)= 14,715N

Then used The equation Ff = u Fn = (0.85)(14,715N) = 12507.75 N which is the force of friction.

Since the force of the applied force and the force of friction are going in opposite directions I subtracted 14,715- 12507.5 and used the Fnet=ma equation getting 1.465 m/s^2.

Would this make sense?

The question:
What is the force of the car?

Last edited: Nov 30, 2011
2. Nov 30, 2011

### Zula110100100

If the applied force were greater than the force of friction your wheels would begin to slip, the problem indicates static friction. So the wheels can't be slipping and the force applied must be equal to or less than the static friction, I believe. Does it give you any info for air-resistance?

3. Nov 30, 2011

### xxharulover

It says to neglect all air-resistance.

4. Nov 30, 2011

### Zula110100100

What do you mean by this exactly?